Construct a confidence interval of the population proportion

Construct a confidence interval of the population proportion at the given level of confidence.

x=2410, n=300, 99% confidence

The 99% cofidence interval is (___,___) (Use ascending order. Round to three decimal places as needed)

Solution

Value of X will always less than size \'n\'. Here it is given as x=2410, n=300. I suppose you mistaken typr 2410 instead of 241 & calculated the respective. And in a second chance also calculated by assuminng n=2410, x=300
a)
CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
x = Mean
n = Sample Size
a = 1 - (Confidence Level/100)
Za/2 = Z-table value
CI = Confidence Interval
Mean(x)=241
Sample Size(n)=300
Sample proportion = x/n =0.8033
Confidence Interval = [ 0.8033 ±Z a/2 ( Sqrt ( 0.8033*0.1967) /300)]
= [ 0.8033 - 2.58* Sqrt(0.0005) , 0.8033 + 2.58* Sqrt(0.0005) ]
= [ 0.7441,0.8625]

Mean(x)=300
Sample Size(n)=2410
Sample proportion = x/n =0.1245
Confidence Interval = [ 0.1245 ±Z a/2 ( Sqrt ( 0.1245*0.8755) /2410)]
= [ 0.1245 - 2.58* Sqrt(0) , 0.1245 + 2.58* Sqrt(0) ]
= [ 0.1071,0.1419]